Answer:
11.8.4 Distillation Columns
Distillation columns present a hazard in that they contain large inventories of flammable boiling liquid, usually under pressure. There are a number of situations which may lead to loss of containment of this liquid.
The conditions of operation of the equipment associated with the distillation column, particularly the reboiler and bottoms pump, are severe, so that failure is more probable.
The reduction of hazard in distillation columns by the limitation of inventory has been discussed above. A distillation column has a large input of heat at the reboiler and a large output at the condenser. If cooling at the condenser is lost, the column may suffer overpressure. It is necessary to protect against this by higher pressure design, relief valves, or HIPS. On the other hand, loss of steam at the reboiler can cause underpressure in the column. On columns operating at or near atmospheric pressure, full vacuum design, vacuum breakers, or inert gas injection is needed for protection. Deposition of flammable materials on packing surfaces has led to many fires on opening of distillation column for maintenance.
Another hazard is overpressure due to heat radiation from fire. Again pressure relief devices are required to provide protection.
The protection of distillation columns is one of the topics treated in detail in codes for pressure relief such as APIRP 521. Likewise, it is one of the principal applications of trip systems.
Another quite different hazard in a distillation column is the ingress of water. The rapid expansion of the water as it flashes to steam can create very damaging overpressures.
Answer:
4Ba(CO3) -> 4BaO2 + 2CO2
Explanation:
I looked at the oxygens to balance this. Ba(CO3) normally has 3 oxygens. BaO2 and CO2 have 4 oxygens total. The common multiple of 3 & 4 is 12. So there should be 12 oxygens on both sides. Then I just found the coefficients that would give 12 oxygens on both sides and can balance the rest of the atoms.
I think the correct answer is C
Answer:
-5.51 kJ/mol
Explanation:
Step 1: Calculate the heat required to heat the water.
We use the following expression.
where,
- c: specific heat capacity
- m: mass
- ΔT: change in the temperature
The average density of water is 1 g/mL, so 75.0 mL ≅ 75.0 g.
Step 2: Calculate the heat released by the methane
According to the law of conservation of energy, the sum of the heat released by the combustion of methane (Qc) and the heat absorbed by the water (Qw) is zero
Qc + Qw = 0
Qc = -Qw = -22.0 kJ
Step 3: Calculate the molar heat of combustion of methane.
The molar mass of methane is 16.04 g/mol. We use this data to find the molar heat of combustion of methane, considering that 22.0 kJ are released by the combustion of 64.00 g of methane.
The answer is <span>Chlorine atoms. This is the </span><span>product of the ultraviolet decomposition of cfcs acts as the catalyst for ozone decomposition. </span>